--- title: "Series field diverter to raise speed to 600 rpm" subject: "Electrical Machines" chapter: 4 key: "semester_4/electrical_machines:dcm-series-500v-diverter-600rpm" source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-series-500v-diverter-600rpm site: BCT Engineering --- # Series field diverter to raise speed to 600 rpm Electrical Machines · Chapter 4: DC Motor - Asked in: 2074 Bhadra (old course) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-series-500v-diverter-600rpm - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4 ## Question A 500 V dc series motor runs at 500 rpm and takes 60 A. The resistance of the field and the armature are $0.3\ \Omega$ and $0.2\ \Omega$ respectively. Calculate the value of the resistance to be shunted with the series field in order that speed be increased to 600 rpm, if the load torque is assumed to be constant. Saturation may be neglected. ## Short answer Using T∝ΦI_a constant and Φ∝I_f, solve: new armature current 71.91 A, diverter resistance = 0.688 Ω. ## Answer Use $\Phi\propto I_f$ (saturation neglected), $T\propto\Phi I_a$ and $E_b\propto\Phi N$. With the diverter in parallel with the field only, the armature current is the line current $I_a$, and the field current is a fraction $k = \dfrac{R_d}{R_d + R_f}$ of it. ```text Ia +----[ Rd ]----+ ---->----[Ra]---+ +----> +---[ Rf ]-----+ If = k.Ia ``` #### Initial condition (no diverter) $$ I_{a1} = I_{f1} = 60\ \text{A}, \qquad E_{b1} = 500 - 60(0.2 + 0.3) = 470\ \text{V} $$ #### New condition (600 rpm, same load torque) Let the new armature current be $I_{a2} = x$ and $I_{f2} = kx$. **Torque constant:** $\Phi_1I_{a1} = \Phi_2I_{a2}$, so $60\times 60 = kx\cdot x$: $$ kx^2 = 3600 \;\Rightarrow\; I_{f2} = kx = \frac{3600}{x} $$ **Back emf:** $$ E_{b2} = 500 - 0.2x - 0.3(kx) = 500 - 0.2x - \frac{1080}{x} $$ **Speed ratio:** $$ \frac{E_{b2}}{E_{b1}} = \frac{\Phi_2N_2}{\Phi_1N_1} = \frac{I_{f2}}{60}\times\frac{600}{500} \;\Rightarrow\; E_{b2} = 470\times 1.2\times\frac{3600/x}{60} = \frac{33840}{x} $$ Equating the two expressions for $E_{b2}$: $$ 500 - 0.2x - \frac{1080}{x} = \frac{33840}{x} \;\Rightarrow\; 0.2x^2 - 500x + 34920 = 0 $$ $$ x = \frac{500 - \sqrt{500^2 - 4(0.2)(34920)}}{2(0.2)} = 71.91\ \text{A} $$ (The smaller root is taken because the motor current stays near the original value.) #### Diverter resistance $$ I_{f2} = \frac{3600}{x} = 50.06\ \text{A}, \qquad k = \frac{I_{f2}}{I_{a2}} = 0.6962 $$ $$ \frac{R_d}{R_d + R_f} = k \;\Rightarrow\; R_d = \frac{kR_f}{1 - k} = \frac{0.6962\times 0.3}{1 - 0.6962} = 0.688\ \Omega $$ Check: $E_{b2} = 500 - 0.2(71.91) - 0.3(50.06) = 470.60\ \text{V}$, and $470\times 1.2\times(50.06/60) = 470.60\ \text{V}$. **Answer:** the diverter resistance is $R_d \approx 0.688\ \Omega$ (the motor then draws about $71.91$ A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A diverter in parallel with the series field does what? A. Lets only a fraction of the line current flow in the field, weakening flux B. Increases flux C. Reduces armature current only D. Shorts the armature Answer: A. Field current becomes k Ia with k = Rd/(Rd+Rf) < 1. (Key point: Diverter placement) 2. With constant load torque, what relation links the two conditions? A. Ia1 = Ia2 B. Phi1 Ia1 = Phi2 Ia2 C. Eb1 = Eb2 D. N1 = N2 Answer: B. T is proportional to Phi Ia and torque is constant. (Key point: Constant torque condition) 3. Initial Eb1 is: A. 500 V B. 490 V C. 480 V D. 470 V Answer: D. 500 - 60(0.2 + 0.3) = 470 V. (Key point: Initial back emf) 4. The diverter resistance found is about: A. 0.3 ohm B. 0.688 ohm C. 1.2 ohm D. 0.2 ohm Answer: B. k = 0.6962, Rd = kRf/(1 - k) = 0.688 ohm. (Key point: Result) ## Related in Electrical Machines - [Field diverter to raise series motor speed](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aseries-motor-500v-diverter-600rpm) — Electrical Machines I, chapter 4 - [Shunt motor: field resistance to raise speed to 1100 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-field-resistance-1100rpm) — Electrical Machines, chapter 4 - [Armature resistance to reduce shunt motor speed to 1250 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-30a-1250rpm-armature) — Electrical Machines I, chapter 4 - [Shunt motor speed with series armature resistance](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-half-full-load-torque) — Electrical Machines, chapter 4 - [Shunt motor speed with 50 Ω field resistance and lower torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-25a-50ohm-torque) — Electrical Machines I, chapter 4, asked 4× - [Shunt motor speed with 50 Ω field resistance (1200 rpm)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-25a-half-load) — Electrical Machines I, chapter 4, asked 5× ## Related in other subjects - [Shunt motor speed with 1.2 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-30a-1p2ohm-torque) — Electrical Machines I, chapter 4, asked 5× - [Series resistance to halve series motor speed](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aseries-motor-600rpm-250v-resistance-300rpm) — Electrical Machines I, chapter 4 - [Series motor resistance for 800 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aseries-motor-armature-resistance-800rpm) — Electrical Machine, chapter 4, asked 2× - [Shunt motor speed with 75 Ω field resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-75ohm-torque-20pc) — Electrical Machines I, chapter 4, asked 2×